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\title{Chapter 08: Noetherian Rings and Modules}
\author{SCC ET AL}
%\institute[XX大学]{XX大学\quad 数学与统计学院\quad 数学与应用数学专业}
%\date{2025年6月}

\begin{document}

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% 封面页
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  \titlepage
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% 目录页
\begin{frame}{Contents}
  \tableofcontents
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% Section 0
%\section{INTRO.}
\begin{frame}[allowframebreaks]{intro. }

\vspace{-0.3cm}

There is very little that one can say about a general ring and its modules. 

In practice an interesting structure theory will result either if the ring has a topology (which is compatible with its operations), or if it has finite dimension, or some generalization thereof. 

As an example of the former, we have the theory of $C^*$-algebras. 

The latter class includes many important rings: algebras that are finite dimensional over a field, PI rings, artinian rings and noetherian rings. 

It is the last ones that we now study. 

In particular, we prove that the Weyl algebra is a noetherian ring.

\end{frame}

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% Section 1
\section{Noetherian Modules}
\begin{frame}[allowframebreaks]{A. }

\vspace{-0.3cm}

In this book we shall be concerned almost exclusively with finitely generated modules. 

One easily checks that a homomorphic image of a finitely generated module is finitely generated. 

However a finitely generated module can have a submodule that is not itself finitely generated. 

An example is the polynomial ring in infinitely many variables $K[x_1,x_2,\ldots]$. 

Taken as a module over itself this ring is a cyclic left module: it is generated by 1. 

However, the ideal generated by all the variables $x_1,x_2,\ldots$ cannot be finitely generated: every finite set of polynomials in $K[x_1,x_2,\ldots]$ uses up only finitely many of the variables.

We get around this problem with a definition. 

A left $R$-module is called {\color{red}noetherian} if all its submodules are finitely generated. 

Examples are easy to come by: vector spaces over $K$ are noetherian $K$-modules. 

Every ideal of the polynomial ring in one variable $K[x]$ is a noetherian $K[x]$-module.

There are several equivalent ways to define noetherianness. 

We chose the most natural. 

Here are two more.

\textbf{Theorem 1.1.} 

Let $M$ be a left $R$-module. 

The following conditions are equivalent:

(1) $M$ is noetherian.

(2) For every infinite ascending chain $N_1 \subseteq N_2 \subseteq \ldots$ of submodules of $M$ there exists $k \geq 0$ such that $N_i = N_k$ for every $i \geq k$.

(3) Every set $\mathcal{S}$ of submodules of $M$ has a submodule $L$ which is not properly contained in any submodule of $\mathcal{S}$.

Condition (2) is known as the {\color{red}ascending chain condition}. 

This is probably the most common way in which noetherian modules are defined. 

Condition (3) is the {\color{red}maximal condition}: the submodule $L$ is called a {\color{red}maximal element} of $\mathcal{S}$.

\textbf{Proof:} Suppose that (1) holds. 

If $N_1 \subseteq N_2 \subseteq \ldots$ is an infinite ascending chain of submodules of $M$, then $Q=\bigcup_{i \geq 1} N_i$ is a submodule of $M$. 

Thus $Q$ is generated by finitely many elements, say $u_1,\ldots,u_t$. 

Hence there exists $k \geq 0$ such that $u_1,\ldots,u_t \in N_k$. 

Thus $Q=N_k=N_i$, for every $i \geq k$, as claimed in (2).

Assume now that (2) holds. 

We prove (3) by reaching a contradiction. 

Suppose that $\mathcal{S}$ does not contain a maximal element. 

If $N_1 \subseteq N_2 \subseteq \cdots \subseteq N_k$ is any chain of elements of $\mathcal{S}$, then we can make it longer. 

Since no element of $\mathcal{S}$ is maximal, there must exist $N_{k+1}$ in $\mathcal{S}$ such that $N_k \subset N_{k+1}$. 

In this way we construct a proper infinite ascending chain of submodules of $M$, thus contradicting (2).

Finally, assume (3) and let $N$ be a submodule of $M$. 

Let $\mathcal{S}$ be the set of all finitely generated submodules of $N$, and let $L$ be a maximal element of $\mathcal{S}$. 

Suppose $L \subset N$, and choose $u \in N - L$.

Thus $L + Ru$ is finitely generated and contains $L$ properly: a contradiction. 

Hence $L = N$, is finitely generated.

In the next proposition we collect the basic properties of noetherian modules that are used in later sections. 

First, a technical lemma.

\textbf{Lemma 1.2.} 

Let $M$ be a left module over a ring $R$. 

Let $N$, $P_1$ and $P_2$ be submodules of $M$ such that $P_2 \subseteq P_1$. 

If $N + P_1 = N + P_2$ and $N \cap P_1 = N \cap P_2$, then $P_1 = P_2$.

\textbf{Proof:} We need only show that $P_1 \subseteq P_2$. 

Suppose that $u_1 \in P_1$. 

Since
\begin{equation}
u_1 \in N + P_1 = N + P_2,
\end{equation}
there exist $v_1 \in N$ and $w_1 \in P_2$ such that $u_1 = v_1 + w_1$. 

Then $v_1 = u_1 - w_1 \in P_1$ since $w_1 \in P_2 \subseteq P_1$. 

Therefore $v_1 \in N \cap P_1 = N \cap P_2$. 

Thus $v_1 \in P_2$. 

Consequently, $u_1 = v_1 + w_1 \in P_2$.

We have that $u_1 = x + u_2$, for $x \in N$ and $u_2 \in P_2$. 

Thus
\begin{equation}
x = u_1 - u_2 \in P_1 \cap N = P_2 \cap N.
\end{equation}

In particular, $x \in P_2$. 

Hence $u_1 = x + u_2 \in P_2$, as required.

\textbf{Proposition 1.3.} 

Let $M$ be a left module over a ring $R$, and let $N$ be a submodule of $M$.

(1) $M$ is noetherian if and only if $M/N$ and $N$ are noetherian.

(2) Let $N'$ be another submodule of $M$ and suppose that $M = N + N'$. 
%
If $N, N'$ are noetherian, then so is $M$.

\textbf{Proof:} It is clear that a submodule of a noetherian module is itself noetherian. 

On the other hand, a submodule of $M/N$ is of the form $L/N$ for some submodule $L$ of $M$ which contains $N$. 

If $M$ is noetherian, then $L$ is finitely generated. 

Thus $L/N$ is finitely generated, and $M/N$ is noetherian.

Conversely, suppose that $N$ and $M/N$ are noetherian. 

Let $L_1 \subseteq L_2 \subseteq \ldots$ be an infinite ascending chain of submodules of $M$. 

Then
\begin{equation}
(L_1 \cap N) \subseteq (L_2 \cap N) \subseteq \ldots
\end{equation}
is an ascending chain of submodules of $N$. 

Since $N$ is noetherian, this chain must stop. 

In other words, there exists $s$ such that $(L_s \cap N) = (L_i \cap N)$ for all $i \geq s$. 

Similarly, the chain
\begin{equation}
N + L_1/N \subseteq N + L_2/N \subseteq \ldots
\end{equation}
stops; that is, there exists $r$ such that $N + L_r = N + L_i$, for $i \geq r$. 

Taking $t = \max\{s, r\}$, we have that $N + L_t = N + L_i$ and $L_t \cap N = L_i \cap N$, for $i \geq t$. 

Thus by Lemma 1.2, $L_t = L_i$ for $i \geq t$. 

Hence $M$ is a noetherian ring.

To prove (2), note that $M/N \cong N'/(N' \cap N)$ is noetherian by (1). 

Since $N$ is noetherian, we may use (1) again to conclude that $M$ is noetherian.

We have pitifully few examples of noetherian modules, but this situation will be rectified in the next section.


\end{frame}

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% Section 2
\section{Noetherian Rings}
\begin{frame}[allowframebreaks]{B. }

\vspace{-0.3cm}

We say that a ring $R$ is a {\color{red}left noetherian ring} if $R$ is noetherian as a left $R$-module. 

Examples are fields and any principal ideal domain, like $\mathbb{Z}$ or $K[x]$. 

Actually, the most important rings in algebraic geometry are noetherian, because the ring of polynomials in finitely many variables is noetherian. 

This was proved by D. Hilbert in 1890 and it is the keystone of commutative algebra. 

It is known as {\color{red}Hilbert's basis theorem}. 

We shall now prove a modernized version of Hilbert's result from which the original theorem follows as an easy consequence.

\textbf{Theorem 2.1.} 

Let $R$ be a commutative noetherian ring. 

The polynomial ring $R[x]$ is noetherian.

\textbf{Proof:} Suppose that $R[x]$ is not noetherian. 

Let $I$ be an ideal of $R[x]$ that is not finitely generated. 

We shall construct, inductively, an infinite ascending chain of ideals in $R$; thereby achieving a contradiction. 

Choose $f_1 \in I$ of smallest possible degree. 

Assume, by induction, that $f_1,\ldots,f_k$ have been chosen. 

Let $f_{k+1}$ be the polynomial of smallest possible degree in $I - (f_1,\ldots,f_k)$. 

Since $I$ is not finitely generated, this construction produces an infinite sequence of polynomials $f_1,f_2,\ldots \in I$. 

Let $n_i$ be the degree and $a_i$ the leading coefficient of $f_i$.

Since $R$ is noetherian, the ascending chain of ideals
\begin{equation}
(a_1) \subseteq (a_1,a_2) \subseteq \ldots
\end{equation}
must be stationary; say $(a_1,\ldots,a_k) = (a_1,\ldots,a_{k+1})$. 

Hence $a_{k+1} = \sum_{i=1}^{k} b_i a_i$, for some $b_i \in R$. 

Put
\begin{equation}
g = f_{k+1} - \sum_{i=1}^{k} b_i x^{n_{k+1}-n_i} f_i.
\end{equation}

Since, by construction, the degrees satisfy $n_1 \leq n_2 \leq \ldots$, it follows that $g$ is indeed a polynomial. 

Note that $g \in I$, but $g \notin (f_1,\ldots,f_k)$. 

However $g$ has smaller degree than $f_{k+1}$, a contradiction. 

Thus $I$ must be finitely generated.

A simple induction using Theorem 2.1 is all that is needed to prove the original basis theorem of Hilbert.

\textbf{Corollary 2.2.} 

Let $K$ be a field. 
%
The polynomial ring $K[x_1,\ldots,x_n]$ is noetherian.

Although every ideal of $K[x_1,\ldots,x_n]$ is finitely generated, it is not true that there is an upper bound on the number of generators of ideals in this ring. 

For example, it is easy to construct ideals $I_k$ of $K[x_1,x_2]$ which cannot be generated by less than $k$ elements; see Exercise 4.2.

We shall now use Hilbert's basis theorem to prove that $A_n$ is a left noetherian ring. 

This will follow from a slightly more general theorem. 

Recall that $\mathcal{B}$ is the Bernstein filtration of $A_n$ and $S_n = \operatorname{gr}^{\mathcal{B}} A_n$.

\textbf{Theorem 2.3.} 

Let $M$ be a left $A_n$-module with a filtration $\Gamma$ with respect to the Bernstein filtration $\mathcal{B}$. 

If $\operatorname{gr}^\Gamma M$ is a noetherian $S_n$-module, then $M$ is noetherian.

\textbf{Proof:} Let $N$ be a submodule of $M$, and $\Gamma'$ the filtration of $N$ induced by $\Gamma$; see Ch.7, §5. 

Since $\operatorname{gr}^{\Gamma'} N \subseteq \operatorname{gr}^\Gamma M$ and the latter is noetherian, we conclude that $\operatorname{gr}^{\Gamma'} N$ is finitely generated.

Since the generators of $\operatorname{gr}^{\Gamma'} N$ are finite in number, they have degree $\leq m$, for some integer $m$. 

We wish to show that $N$ is generated by the elements in $\Gamma'_m$. 

Suppose that it is not, and let $k$ be the smallest integer for which there exists $v \in \Gamma'_k$ which cannot be generated by the elements of $\Gamma'_m$. 

Clearly $k > m$. 

Let $\mu_k$ be the symbol map of order $k$ of $\Gamma'$. 

By the hypothesis on $\operatorname{gr}^{\Gamma'} N$, there exist $a_i \in A_n$ and $u_i \in \Gamma'_{r_i}$ such that
\begin{equation}
\mu_k(v) = \sum_{i=1}^{s} \sigma_{k-r_i}(a_i) \mu_{r_i}(u_i)
\end{equation}
where $r_i \leq m$, for $i=1,2,\ldots,s$. 

Hence
\begin{equation}
v - \sum_{i=1}^{s} a_i u_i \in \Gamma'_{k-1},
\end{equation}
which, by the minimality of $k$, may be written as an $A_n$-linear combination of elements in $\Gamma'_m$. 

Thus $v$ itself may be written as an $A_n$-linear combination of elements in $\Gamma'_m$.

To finish the proof we must show that a finite number of elements of $\Gamma'_m$ are enough to generate $N$. 

But $\Gamma'_m$ is a subspace of the finite dimensional vector space $\Gamma_m$. 

Hence it has a finite basis, which will be a set of generators for $N$ as an $A_n$-module.


\textbf{Corollary 2.4.} 

$A_n$ is a left noetherian ring.

\textbf{Proof:} The graded ring $S_n$ associated to the Bernstein filtration is a polynomial ring in $2n$ variables by Theorem 7.3.1. 

Hence, by Hilbert's basis theorem, $S_n$ is noetherian. 

Thus $A_n$ is left noetherian by Theorem 2.3.

It has already been pointed out in Ch.2 that every left ideal of $A_n$ can be generated by two elements. 

Therefore, there is an upper bound on the number of generators of left ideals in $A_n$, which is not the case for a polynomial ring. 

However, this result is very hard to prove and is beyond our means in this book.

There are good reasons why one ought to rejoice that $A_n$ is a left noetherian ring. 

For example, it follows from Corollary 2.4 and the next proposition that every finitely generated $A_n$-module is noetherian.

\textbf{Proposition 2.5.} 

Let $R$ be a left noetherian ring. 

Finitely generated left $R$-modules are noetherian.

\textbf{Proof:} Let $M$ be a finitely generated left $R$-module. 

If $M$ is generated by $k$ elements then there exists a surjective homomorphism $\phi: R^k \longrightarrow M$. 

Since $R$ is left noetherian, it follows from Proposition 1.3(2) that $R^k$ is noetherian. 

Hence $M$ is noetherian by 1.3(1).

There is really nothing special about left modules in this context, and all the results that we have stated so far hold for right modules with the obvious changes. Thus the ring $A_n$ is also right noetherian, and we shall use this in later chapters, without further comment.


\end{frame}

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% Section 3
\section{Good Filtrations}
\begin{frame}[allowframebreaks]{C. }

\vspace{-0.3cm}

Let $M$ be a left $A_n$-module and $\Gamma$ a filtration of $M$ with respect to the Bernstein filtration $\mathcal{B}$. 

If $\operatorname{gr}^\Gamma M$ is finitely generated, then it is noetherian by Proposition 2.5. 

Hence $M$ is finitely generated, by Theorem 2.3. 

However, it is not always true that if $M$ is finitely generated over $A_n$ then $\operatorname{gr}^\Gamma M$ is finitely generated over $S_n$. 

When $\operatorname{gr}^\Gamma M$ is finitely generated we say that $\Gamma$ is a {\color{red}good} filtration of $M$.

It is nonetheless true that every finitely generated $A_n$-module admits a good filtration. 

Indeed, if $M$ is generated by $u_1,\ldots,u_s$ then the filtration
\begin{equation}
\Gamma_m = \left\{\sum_{i=1}^{s} a_i u_i \mid a_i \in \mathcal{B}_m\right\}
\end{equation}
is a good filtration of $M$. 

To see this, note that $\operatorname{gr}^\Gamma M$ is generated by the elements $\mu_k(u_i)$ where $k$ is the degree of $u_i$ in $\Gamma$. 

Thus $\operatorname{gr}^\Gamma M$ is finitely generated over $S_n$.


$\Gamma$ of $M$ defined by $\Gamma_k = \sum_{i}^{s} B_k u_i$ is good. 

The graded module $\operatorname{gr}^\Gamma M$ is generated over $S_n$ by the symbols of $u_1,\ldots,u_s$.

We need an example of a filtration that is not good. 

Before that, however, we shall establish an easy criterion to check whether a filtration is good.

\textbf{Proposition 3.1.} 

Let $M$ be a left $A_n$-module. 

A filtration $\Gamma$ of $M$ with respect to $\mathcal{B}$ is good if and only if there exists $k_0$ such that $\Gamma_{i+k} = B_i \Gamma_k$, for all $k \geq k_0$.

\textbf{Proof:} Suppose that there exists $k_0$ such that $\Gamma_{i+k} = B_i \Gamma_k$, for all $k \geq k_0$. 

The $K$-vector space $\Gamma_{k_0}$ has a finite basis. 

The symbols of the elements in this basis generate $\operatorname{gr}^\Gamma M$. 

Thus $\Gamma$ is good.

Conversely, suppose that $\operatorname{gr}^\Gamma M$ is finitely generated over $S_n$. 

Let $u_1,\ldots,u_s$ be elements of $M$ whose symbols generate $\operatorname{gr}^\Gamma M$. 

Assume that $u_j \in \Gamma_{k_j} - \Gamma_{k_j-1}$, for $j=1,2,\ldots,s$, and that $k_0 = \max\{k_1,\ldots,k_s\}$.

Let $k \geq k_0$. 

We prove that $\Gamma_{i+k} = B_i \Gamma_k$ by induction on $i$. 

If $i=0$, the result is obviously true. 

Suppose that the equality holds for $i-1$. 

Pick $v \in \Gamma_{i+k}$. 

Since $\operatorname{gr}^\Gamma M$ is finitely generated, and denoting by $\mu_k$ the symbol of order $k$ of $\Gamma$, one has
\begin{equation}
\mu_{i+k}(v) \in \sum_{j=1}^{s} \sigma_{k+i-k_j}(B_{k+i-k_j}) \mu_{k_j}(u_j).
\end{equation}

Since $B_{i+k-k_j} = B_i \cdot B_{k-k_j}$, we conclude that

\begin{equation}
v \in \sum_{j=1}^{s} B_i \cdot B_{k-k_j} u_j + \Gamma_{i+k-1}.
\end{equation}

Now $B_{k-k_j} u_j \in \Gamma_k$ for every $j$ and, by the induction hypothesis, $\Gamma_{k+i-1} = B_{i-1} \Gamma_k$. 

Since $B_{i-1} \subseteq B_i$, then $v \in B_i \Gamma_k$. 

Thus $\Gamma_{i+k} \subseteq B_i \Gamma_k$. 

The other inclusion is obvious; hence we have an equality, as desired.

It is now easy to check whether a filtration is good or not. 

For example, let $\Omega$ be the filtration of the left module $A_n$ defined by $\Omega_k = B_{2k}$. 

Then we have that $B_i \Omega_k = B_{i+2k}$ is properly contained in $B_{2(i+k)} = \Omega_{i+k}$. 

By Proposition 3.1, $\Omega$ is not a good filtration of $A_n$ with respect to $\mathcal{B}$.

We end this section with a proposition that will allow us to compare two good filtrations.


\textbf{Proposition 3.2.} 

Let $M$ be a left $A_n$-module. 

Suppose that $\Gamma$ and $\Omega$ are two filtrations of $M$ with respect to $\mathcal{B}$.

(1) If $\Gamma$ is good then there exists $k_1$ such that $\Gamma_j \subseteq \Omega_{j+k_1}$.

(2) If $\Gamma$ and $\Omega$ are good then there exists $k_2$ such that $\Omega_{j-k_2} \subseteq \Gamma_j \subseteq \Omega_{j+k_2}$.

\textbf{Proof:} By Proposition 3.1 there exists $k_0$ such that $B_i \Gamma_j = \Gamma_{i+j}$, for all $j \geq k_0$ and $i \geq 0$. 

Since $\Gamma_{k_0}$ is a finite dimensional vector space over $K$, there exists $k_1$ such that $\Gamma_{k_0} \subseteq \Omega_{k_1}$. 

Thus $\Gamma_{j+k_0} = B_j \Gamma_{k_0}$ is contained in $B_j \cdot \Omega_{k_1} \subseteq \Omega_{j+k_1}$. 

Hence
\begin{equation}
\Gamma_j \subseteq \Gamma_{j+k_0} \subseteq \Omega_{j+k_1}
\end{equation}
for all $j \geq 0$, which proves (1). 

To prove (2) apply (1) twice, swapping $\Gamma$ and $\Omega$.

The full power of good filtrations will only be felt in the next chapter, where they will be used to define a dimension for finitely generated $A_n$-modules.


\end{frame}

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% Section 4
\section{Exercises}
\begin{frame}[allowframebreaks]{D. }

\vspace{-0.5cm}

\textbf{Exercise 4.1.} A $K$-algebra $R$ is {\color{red}affine} if there exist elements $r_1,\ldots,r_s \in R$ so that the monomials $r_1^{m_1} \cdots r_s^{m_s}$ form a $K$-basis for $R$.

(1) Show that if $R$ is an affine commutative $K$-algebra then it is a homomorphic image of a polynomial ring in finitely many variables over $K$.

(2) Use (1) and Hilbert's basis theorem to conclude that a commutative affine $K$-algebra is noetherian.

(3) Show that (2) is false if the commutative hypothesis is dropped.

%Hint: Construct an infinite ascending chain of left ideals for the free algebra $K\{x,y\}$.


\newpage 

\textbf{Exercise 4.2.} Let $I_k$ be the ideal of the polynomial ring $K[x_1,x_2]$ generated by the monomials of degree $k$. 
%
In other words, the generators of $I_k$ are of the form $x_1^i x_2^j$ with $i+j=k$. 

Let $\mathcal{M}=I_1$.

(1) Show that $\mathcal{M}$ is a maximal ideal of $K[x_1,x_2]$, and that $K[x_1,x_2]/\mathcal{M} \cong K$.

(2) Show that $I_k/\mathcal{M} I_k$ is a $K$-vector space of dimension $k+1$.

(3) Show that $I_k$ cannot be generated by less than $k+1$ elements.


\newpage 

\textbf{Exercise 4.3.} Show that the results of \S 3 remain true if we replace the Bernstein filtration $\mathcal{B}$ by the filtration by order.

Hint: In some of the proofs it will be necessary to use Hilbert's basis theorem and Proposition 2.5.


\newpage 

\textbf{Exercise 4.4.} Let $M$ be a left $A_n$-module with a filtration $\Gamma$ with respect to $\mathcal{B}$. 

Suppose that $u_i \in \Gamma_{k_i}$, for $i=1,\ldots,s$. 

Show that if $\operatorname{gr}^\Gamma M$ is generated by $\mu_{k_i}(u_i)$, for $i=1,\ldots,s$, then $M$ is generated by $u_1,\ldots,u_s$.


\newpage 

\textbf{Exercise 4.5.} The purpose of this exercise is to show that the converse of Exercise 4.4 is false. 
%
Let $J$ be the left ideal of $A_3$ generated by $\partial_1$ and $\partial_2 + x_1 \partial_3$. 

Recall that by Exercise 6.7.7, if $\mathcal{B}'$ is the filtration of $J$ induced by the Bernstein filtration of $A_3$, then $\operatorname{gr}^{\mathcal{B}'} J \cong \operatorname{gr} J$.

(1) Using Exercise 2.4.2, show that $\operatorname{gr} J$ is generated by $\sigma_1(\partial_i)$ for $i=1,2,3$.

(2) Show that although $\partial_1$ and $\partial_2 + x_1 \partial_3$ generate $J$, their symbols do not generate $\operatorname{gr} J$.

\newpage 

\textbf{Exercise 4.6.} Let $M$ be a left $A_n$-module with a good filtration $\Gamma$. 

Show that the annihilator of $\operatorname{gr}^\Gamma M$ is a homogeneous ideal of $S_n$.




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